*sigh*
Okay, marcus. You made me do it. (You have the right idea.) Let's talk about how diodes (and, indirectly, any discontinuous materials that exhibit a contact potential, like thermocouples) work. I do not have my preferred semiconductor device physics textbook with me at the moment (Neaman), so I'm running from memory and might make some mistakes.
First, let's describe the difference, in microscopic (quantum-mechanical) terms, between conductors and insulators. In an insulator, electrons are rather strongly bound to individual atoms, and cannot move around much. In a metal, some of the electrons are free to wander around the bulk material. In a sense, the electrons form a "free electron gas" inside the material. It is these free electrons which can carry electrical current.
Electrons which are bound to atoms are called valence electrons; electrons which are free to move about are called conduction electrons.
Classically, an electron could have any energy in a material; but quantum-mechanically, this is not so. The density of allowed electron states is not uniform. There is, in fact, a gap between two classes of states: valence and conduction. There are many valence states, and many conduction states, but none in between. The near-continuum of valence states is called the 'valence band,' while the near-continuum of conduction states is called the 'conduction band.' The gap between the valence and conduction bands is called the 'bandgap.'
In a conductor, there are always electrons present in the conduction band. In an insulator, there are never electrons present in the conduction band. Semiconductors straddle the fence and can essentially behave as either conductors or insulators -- hence their great utility in electronic switches.
Ah, I found a picture:
http://ece-www.colorado.edu/~bart/book/book/chapter2/ch2_3.htm#2_3_3_1
Take a look at Figures 2.3.4 and 2.3.6 on this page. (The rest of the page is good, too!) Figure 2.3.4 shows a diagram of the states in a semiconductor. Energy is plotted upwards on the figure, and the horizontal direction is, well, rather meaningless, but people usually think of it as spatial.
In any event, you can see the regions labelled: on the bottom there is the valence band, which is mostly full (a band shaded with red would be full, one left white would be empty, in this schematic representation). The highest energy state in the valence band is labelled with energy E
v. On top is the conduction band. The lowest energy state is labelled with energy E
c. In between is the bandgap -- the no man's land. The difference between E
c and E
v is the "size" of the bandgap in units of energy, and is labelled E
g.
This author has also chosen to always include the work function of the metal and the implied vacuum energy, but we're not going to worry about any of that.
Figure 2.3.6 shows the difference, in microscopic terms, between various sorts of materials.
There is a way to describe how full or how empty the bands are. Electrons are fermions, and thus follow Fermi-Dirac statistics (I apologize for the laymen who might not understand what this means -- I don't have the space to explain it here).
Fermi-Dirac statistics:
http://scienceworld.wolfram.com/physics/Fermi-DiracDistribution.html
The "Fermi function" n(E) is a distribution function which describes the percentage of particles in a particular energy state E at a particular temperature T. I'm going to eliminate the terms for the number of particles and express the energy E as the difference from some parameter, E
F, so that the Fermi function looks like:
f(E) = 1 / (1 + exp[ (E-E
F)/kT ])
What is this parameter E
F? Inspection yields that when E = E
F, f(E) is 1/2. The Fermi energy, E
F, is a specially chosen value of energy such that exactly half of the particles are in states of higher energy and exactly half of the particles are in states of lower energy. The Fermi energy E
F is then, in qualitative terms, the "midpoint" energy of the system.
So what does this Fermi energy matter? I've introduced it only because it has this special property of being right in the middle of the number of states. We use the Fermi energy to describe semiconductors, of course.
In a semiconductor, the Fermi energy is always somewhere in the bandgap. Of course, no particles can have energies in the bandgap, so no particles have E = E
F, but that doesn't stop E
F from doing its job for us: half of the electrons in the material have energies higher than E
F, and half have energies lower than it. If the Fermi energy is in the bandgap, then, it's clear that there have to be particles in the conduction band! In fact, if the Fermi energy is
exactly in the middle of the bandgap, then exactly half of the electrons are in the conduction band. This is what happens in a normal, "native," or "intrinsic" semiconductor -- a big ol' block of crystalline silicon will look this way.
But a diode is not just piece of silicon -- it's a piece of silicon that some engineers screwed with in a particular way. The engineers took two pieces of silicon, and subjected each to a hot gas of substances like boron or phosphorus, so that the crystalline silicon absorbs some of the substance. Here and there, the lattice of silicon atoms is disrupted -- there's a boron or phosphorus atom sitting where there used to be a silicon atom. These "impurity" atoms add free electrons to the material, or provide sites where electrons become tightly bound. When you dope silicon with phosphorus, you get a material with an excess of electrons -- it's called an 'n-type' (think negative) semiconductor. When you dope silicon with boron, you get a material with a scarcity of electrons -- it's called a 'p-type' (think positive) semiconductor. We can also think of the scarcity of electrons as an abundance of 'holes,' and consider the holes to be positively charged particles, but that's of no matter.
So now you've got one piece of silicon with too many free electrons, and one with two few. What happens to the Fermi energies in these two materials? Well, the n-type semiconductor is more like a conductor, so its Fermi energy moves up, toward the conduction band. This is just a way of saying that more electrons are present in the conduction band. Similarly, the Fermi energy moves downward in the p-type material, toward the valence band.
So let's do something evil! Let's stick our two pieces of silicon together and make them fight!
If you stick a piece of p-type and n-type semiconductor together, they do fight, but not for long. There are too many electrons in the n-type piece, and too few in the p-type piece. So the electrons in the n-type jump ship and run to the greener pastures of the p-type. But this process can only go on for so long -- as the electrons move, they're building up an electric field which discourages further movement. So only
some electrons move. When you put n-type and p-type together, they do
not totally neutralize each other -- only in a very small region near the contact boundary.
The region of neutralization is called a 'depletion region,' because there are no carriers there -- the semiconductor is neutral. The electric field created by those electrons that jumped ship creates a potential difference between the p-type and n-type semiconductor slabs, and this potential is called, appropriately, a contact potential. Yes, when you put make a "pn" junction, there suddenly exists a voltage across the junction. (This doesn't mean, of course, you've invented a battery.)
Now have a look at
http://ece-www.colorado.edu/~bart/book/book/chapter4/ch4_2.htm
Figure 4.2.3 shows an energy-band diagram of a pn junction after all the electrons have jumped ship -- one would say "after it has reached thermal equilibrium."
The important things to notice are the following:
1) The p-type semiconductor is at a higher electrical potential.
2) The Fermi energy is everywhere constant across the junction.
3) There is an energy barrier for charged carriers to cross the junction. The energy barrier is the contact potential.
Now, when you apply an external voltage to the pn junction, what happens? There are two possibilities. If you apply a positive voltage to the p-side, you
decrease the potential across the junction, and suddenly electrons can start crossing it -- you've turned the diode on.
If you apply a negative voltage to the p-side, you
increase the potential across the junction, and make it even harder for electrons to cross -- you've turned it off, and made its depletion zone even larger.
Take a look at Figure 4.2.4 on
http://ece-www.colorado.edu/~bart/book/book/chapter4/ch4_2.htm for a picture of these conditions.
You can see immediately why diodes turn on at a specific positive voltage now -- usually around 0.6 to 0.7 volts.
The contact potential is about 0.6 to 0.7V!
I hope this helps you understand how diodes work -- and you can probably see why your original idea is flawed.
Oh, and the online book I found is at:
http://ece-www.colorado.edu/~bart/book/book/contents.htm
- Warren
